Can you solve the volume of a cube with unequal heights?

I have a challenge for someone (which I plan on working on this weekend myself when I have free time from my homework).

Can you calculate the volume of a cube-like shape with four different heights and with perfect square base if you are given the surface area of all sides, including the base, except for its top? The sides of this object would be four trapezoids of which two, and only two, heights each side would be equal (where the two trapezoids meet to create a corner). The angle of the trapezoid to its neighbor and the square base is always 90 degrees.

To get a better idea of what I am taking about, check out the picture I uploaded to this post.

I would be very grateful to whomever can solve this for me. Knowing a formula for this would be very useful for a school project I am creating.

In case you were wondering why I want to know this (for motivational purposes), I am trying to map the the volume of space shaded by irregular objects with a light source coming from a given angle (so I can ask an interesting sunlight competition question for my undergraduate senior project in plant ecology). From what I can tell at this point, field measurements could yield an average volume that would be represented by a shape similar to the one I've described (because I believe I've already figured out a way to find the area of all the sides and the base).

I was just hoping a solution already existed for this (so I don't have to reinvent the wheel). Thanks for taking the time to check out my question.

No, this is not technically homework and it probably doesn't belong in this section (this is my first post so I am new here). I am trying to map the the volume of space shaded by irregular objects with a light source coming from a given angle (so I can ask an interesting sunlight competition question for my undergraduate senior project in plant ecology). From what I can tell at this point, field measurements could yield an average volume that would be represented by a shape similar to the one I've described. I was just hoping a solution already existed for this (so I don't have to reinvent the wheel).

Re: Can you solve the volume of "Trapezoidal Prism" with unequal sides?

Your figure appears to have the 4 top points all in the same plane, but I THINK (could be wrong) that this is inconsistent with your statement of the problem which seems to imply that the 4 heights would all be unrelated. Once you pick 3 of the heights, the 4th one is fixed if you want the top to be one plane as you have drawn.

The volume of the whole solid will be (area of base) x (average height). where "average" means (sum of the heights at the four corners)/4.

Even if all 4 top points are in the same plane, that's going to be a VERY messy and tedius set of calculations and if the 4 points are NOT in the same plane then it doesn't work at all (although a similar, but more complicated one, would).

Conceptually of course, it DOES get you to the answer, but surely there's an easier way?

Staff: Mentor

Re: Can you solve the volume of "a cube" with unequal heights?

ThunderSkunk,
It is against forum rules to post the same topic in multiple sections. I am not issuing an warning or infraction this time, as I believe you were acting on the advice of people responding in this thread. I have merged the two threads into this one.

In the future, if you find that you have posted something in the wrong section, click the Report button, and a mentor will take care of moving the thread.

I have a challenge for someone (which I plan on working on this weekend myself when I have free time from my homework).

Can you calculate the volume of a cube-like shape with four different heights and with perfect square base if you are given the surface area of all sides, including the base, except for its top? The sides of this object would be four trapezoids of which two, and only two, heights each side would be equal (where the two trapezoids meet to create a corner). The angle of the trapezoid to its neighbor and the square base is always 90 degrees.

Even if all 4 top points are in the same plane, that's going to be a VERY messy and tedius set of calculations and if the 4 points are NOT in the same plane then it doesn't work at all (although a similar, but more complicated one, would).

Conceptually of course, it DOES get you to the answer, but surely there's an easier way?

V= [itex]\frac{1}{6}[/itex] (h[itex]_{1}[/itex]+2h[itex]_{2}[/itex]+2h[itex]_{3}[/itex] +h[itex]_{4}[/itex]) a[itex]^{2}[/itex] cubic units, where the h[itex]_{4}[/itex] corresponds to the height of the 4th point which is not in the plane of other 3 points , i hope this will also work if the four points are in the same plane . can anybody check it ?

V= [itex]\frac{1}{6}[/itex] (h[itex]_{1}[/itex]+2h[itex]_{2}[/itex]+2h[itex]_{3}[/itex] +h[itex]_{4}[/itex]) a[itex]^{2}[/itex] cubic units, where the h[itex]_{4}[/itex] corresponds to the height of the 4th point which is not in the plane of other 3 points , i hope this will also work if the four points are in the same plane . can anybody check it ?

I have no idea where that equation comes from (and don't know whether it's right or wrong) BUT I find it unlikely that it could be correct since you have a factor of 2 next to 2 of the legs but not next to the other 2. Intuitively it would seem impossible that this could be a correct method.

Just look at a simple degenerative case where all for upper points are at the same height. Now you increase one of the points that you have with a factor of 2 and do the computation. OR you increase one of the points for which you do NOT have a factor of 2, and by the same amount, then do the computation. Clearly you'll get two different answers for the identical figure.

Wow, DaveC426913! That's brilliant! Thank you all very much! I am very surprised that such an elegant solution was lurking behind this problem. I wonder how long it would have taken me to solve this myself... For me this was certainly not trivial. This will be very helpful. Thank you again.

Also, I am very surprised by how many people on this site were willing to help me out on this. Thank all of you for taking the time to look this over for me.

As for Mark44, sorry about violating the site's rules. I'll give those rules a read and make sure I am not trending on anybody's toes from now on.

I think I have solution for the case where the points don't lie along a plane. Critique is welcome...

The approach is to take a series of thin slices from the figure and sum their areas. It is trivial to calculate the area of a slice since the heights of each side (call them p and q) clearly lie upon a line and so we can just take [itex]A = base*(p+q)/2[/itex]. Summing the areas gives the volume of the figure...
[tex]V = \int_0^b bh(x)\,\mathrm{d}x[/tex]
Where b is the length of the base and h(x) is a function giving the average height of each slice, and we sum through the entire figure (from 0 to b).

To derive the height function, let u1 and u2 denote the heights of the near and far corners of the left side of the cube; similarly, use v1 and v2 for the right. We can derive the heights of each side at any given slice by passing a straight line p through points u1,u2 and a line q through v1,v2...
[tex]p = \frac{u1-u2}{b} x + u1 \qquad q = \frac{v1-v2}{b} x + v1[/tex]
We then have...
[tex]h(x) = \frac{\frac{u1-u2}{b} x + \frac{v1-v2}{b} x + (v1 + u1)}{2}[/tex]
Finally giving us...
[tex]V = b\frac{1}{2}\int_0^b {\frac{u1-u2}{b} x + \frac{v1-v2}{b} x + (v1 + u1)}\,\mathrm{d}x[/tex]

The approach seems right, but I may have made a stupid error somewhere along the way. Thoughts?

To derive the height function, let u1 and u2 denote the heights of the near and far corners of the left side of the cube; similarly, use v1 and v2 for the right. We can derive the heights of each side at any given slice by passing a straight line p through points u1,u2 and a line q through v1,v2...

Can we take that approach of a straight line , as if the surface is not a plane then there is a discontinuity from one corner to the opposite corner. There would have either a peak or a valley depending upon if the fourth corner is above or below the plane. I see in my minds eye a planar surface with a wedge added or missing.

I did not rigourously go through your derivation, but is the discontinuity taken into account?

Can we take that approach of a straight line , as if the surface is not a plane then there is a discontinuity from one corner to the opposite corner. There would have either a peak or a valley depending upon if the fourth corner is above or below the plane. I see in my minds eye a planar surface with a wedge added or missing.

I did not rigourously go through your derivation, but is the discontinuity taken into acoount?

Yeah, it doesn't work if the top surface is not a plane. Each slice would be a five-sided shape, possibly convex, possibly concave.

Then again, if it is a plane, using calculus would be using a bulldozer to excavate ... a ... tea ... cup*.